Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $k = \dfrac{n + 5}{5n - 50} \div \dfrac{n^3 - 11n^2 + 10n}{2n^3 - 22n^2 + 20n} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $k = \dfrac{n + 5}{5n - 50} \times \dfrac{2n^3 - 22n^2 + 20n}{n^3 - 11n^2 + 10n} $ First factor out any common factors. $k = \dfrac{n + 5}{5(n - 10)} \times \dfrac{2n(n^2 - 11n + 10)}{n(n^2 - 11n + 10)} $ Then factor the quadratic expressions. $k = \dfrac {n + 5} {5(n - 10)} \times \dfrac {2n(n - 1)(n - 10)} {n(n - 1)(n - 10)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac {(n + 5) \times 2n(n - 1)(n - 10) } {5(n - 10) \times n(n - 1)(n - 10) } $ $k = \dfrac {2n(n - 1)(n - 10)(n + 5)} {5n(n - 1)(n - 10)(n - 10)} $ Notice that $(n - 1)$ and $(n - 10)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac {2n\cancel{(n - 1)}(n - 10)(n + 5)} {5n\cancel{(n - 1)}(n - 10)(n - 10)} $ We are dividing by $n - 1$ , so $n - 1 \neq 0$ Therefore, $n \neq 1$ $k = \dfrac {2n\cancel{(n - 1)}\cancel{(n - 10)}(n + 5)} {5n\cancel{(n - 1)}(n - 10)\cancel{(n - 10)}} $ We are dividing by $n - 10$ , so $n - 10 \neq 0$ Therefore, $n \neq 10$ $k = \dfrac {2n(n + 5)} {5n(n - 10)} $ $ k = \dfrac{2(n + 5)}{5(n - 10)}; n \neq 1; n \neq 10 $